Turn on thread page Beta
    • Thread Starter
    Offline

    21
    ReputationRep:
    Name:  D67609AB-F8EE-4266-B5DC-CC6DD823413C.jpg
Views: 11
Size:  497.5 KB

    This question (part ii) is really annoying me!! Because I almost have the answer but can’t get to the final one and seem to have done everything right…

    I started off with y=mx+c, found xsin2x = 0 at 0,π& 2π, clearly the graph shows P is the first root after 0 so it’s π, then I subbed 2x=π (x=π/2) into tan2x + 2x to get my gradient which again was π, then I wrote that in the form of y=π(x) + c, subbed in P(π/2,0) and got y=π(x)-(π^2/2), so c=π^2/2 when I multiply the equation by 2 I get 2y=2πx-π^2… but they want it in the form of 2πx + 2y = π^2, I cant rearrange my answer into that form…

    I’d really appreciate it if someone could show me where I’m going wrong, It could just be a sign error somewhere in my working
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by Zxyn)

    This question (part ii) is really annoying me!! Because I almost have the answer but can’t get to the final one and seem to have done everything right…

    I started off with y=mx+c, found xsin2x = 0 at 0,π& 2π, clearly the graph shows P is the first root after 0 so it’s π, then I subbed 2x=π (x=π/2) into tan2x + 2x to get my gradient which again was π, then I wrote that in the form of y=π(x) + c, subbed in P(π/2,0) and got y=π(x)-(π^2/2), so c=π^2/2 when I multiply the equation by 2 I get 2y=2πx-π^2… but they want it in the form of 2πx + 2y = π^2, I cant rearrange my answer into that form…

    I’d really appreciate it if someone could show me where I’m going wrong, It could just be a sign error somewhere in my working
    \tan(2x)+2x=0 isn't the gradient function, it's just an equation which is satisfied when you plug in the coordinates of the turning points.
    • Thread Starter
    Offline

    21
    ReputationRep:
    (Original post by RDKGames)
    \tan(2x)+2x=0 isn't the gradient function, it's just an equation which is satisfied when you plug in the coordinates of the turning points.
    Ohhhhh thanks!!! I tried it with the gradient function (sin2x + 2xcos2x) and got -π which fixes everything!

    Thanks for your help
 
 
 
Poll
Favourite type of bread
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.